Chemistry, 16.12.2019 18:31 edwardo9

Alab scientist cools a liquid sample of water (2.6 kg) at 0.00°c to -192°c. the water turns to ice as this temperature change occurs. how much heat is released during this process? [for water, lf = 334 kj/kg and lv = 2257 kj/kg. the specific heat for ice is 2050 j/(kg·k)]. report your answer in kj. (round your answer to a whole number - no decimal places)

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Answer from: carlosiscr7

The heat released for the given process is -1892 kJ

Explanation:

The processes involved in the given problem are:

$1.)H_2O(l)(0^oC,273K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(s)(-192^oC,81K)$

Pressure is taken as constant.

To calculate the amount of heat released at same temperature, we use the equation:

$q=m\times L_{f,v}$ ......(1)

where,

q = amount of heat released = ?

m = mass of water/ice

$L_{f,v}$ = latent heat of fusion or vaporization

To calculate the amount of heat released at different temperature, we use the equation:

$q=m\times C_{p,m}\times (T_{2}-T_{1})$ .......(1)

where,

q = amount of heat released = ?

$C_{p,m}$ = specific heat capacity of medium

m = mass of water/ice

T_2 = final temperature

T_1 = initial temperature

Calculating the heat absorbed for each process:

For process 1:

Converting the latent heat of fusion in J/kg, we use the conversion factor:

1 kJ = 1000 J

So, $(\frac{-334kJ}{1kg})\times (\frac{1000J}{1kJ})=-334\times 10^3J/kg$

We are given:

$m=2.6kg\\L_f=-334\times 10^3J/kg$

Putting values in equation 1, we get:

$q_1=2.6kg\times (-334\times 10^3J/kg)=-868400J$

For process 2:

We are given:

$m=2.6kg\\C_{p,s}=2050J/kg.K\\T_1=273K\T_2=81K$

Putting values in equation 2, we get:

$q_2=2.6kg\times 2050J/kg.K\times (81-(273))^oC\\\\q_2=1023360J$

Total heat absorbed = q_1+q_2

Total heat absorbed = $[-868400+(-1023360)]J=-1891760J=-1891.76kJ\approx -1892kJ$

Hence, the heat released for the given process is -1892 kJ

Answer from: Quest

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

Answer from: Quest

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

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Alab scientist cools a liquid sample of water (2.6 kg) at 0.00°c to -192°c. the water turns to ice a...

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